Question: Find $\lim_{x\to \infty}(2+e^{x})^{^{ \frac1x}}$. Choose 1 answer: Choose 1 answer: (Choice A) A $0$ (Choice B) B $e$ (Choice C) C $\sqrt{e}$ (Choice D) D The limit doesn't exist.
Answer: Taking $x$ to $\infty$ in $(2+e^{x})^{^{ \frac1x}}$ results in the indeterminate form $\infty^{^{0}}$. To make the expression easier to analyze, let's take its natural log (this is a common trick when dealing with composite exponential functions). In other words, letting $y=(2+e^{x})^{^{ \frac1x}}$, we will find $\lim_{x\to \infty}\ln(y)$. Once we find it, we will be able to find $\lim_{x\to \infty}y$. $\ln(y) =\dfrac{\ln(2+e^{x})}{x}$ Taking $x$ to $\infty$ in $\dfrac{\ln(2+e^{x})}{x}$ results in the indeterminate form $\dfrac{\infty}{\infty}$, so now it's L'Hôpital's rule's turn to help us with our quest! $\begin{aligned} &\phantom{=}\lim_{x\to \infty}\ln(y) \\\\ &=\lim_{x\to \infty}\dfrac{\ln(2+e^{x})}{x} \\\\ &=\lim_{x\to \infty}\dfrac{\dfrac{d}{dx}[\ln(2+e^{x})]}{\dfrac{d}{dx}[x]} \gray{\text{L'Hôpital's rule}} \\\\ &=\lim_{x\to \infty}\dfrac{\left(\dfrac{e^{x}}{2+e^{x}}\right)}{1} \\\\ &=\lim_{x\to \infty}\dfrac{e^{x}}{2+e^{x}} \\\\ &=1 \gray{\text{The leading term is the same}} \end{aligned}$ Note that we were only able to use L'Hôpital's rule because the limit $\lim_{x\to \infty}\dfrac{\dfrac{d}{dx}[\ln(2+e^{x})]}{\dfrac{d}{dx}[x]}$ actually exists. We found that $\lim_{x\to \infty}\ln(y)=1$, which means $\lim_{x\to \infty}y=e$. [Why?] In conclusion, $\lim_{x\to \infty}(2+e^x)^{^{\frac{1}{x}}}=e$.